Answer to Question #257023 in Real Analysis for udith

Question

Use the Taylor series to find the values of ln 0.4 accurate to 10^-3 . Use the integral remainder.

Answer 1

Solution:

Using Taylor series expansion:

$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-...$

Put $x=-0.6$

$\ln(1-0.6)=(-0.6)-\dfrac{(-0.6)^2}{2}+\dfrac{(-0.6)^3}{3}-\dfrac{(-0.6)^4}{4}+\dfrac{(-0.6)^5}{5}-... \\ \Rightarrow \ln 0.4\approx-0.6-\dfrac{0.36}{2}-\dfrac{0.216}{3}-\dfrac{0.1296}{4}-\dfrac{0.07776}{5} \\=-0.899952$

$\approx -0.900$ [nearest to $10^{-3}$ or 0.001]